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Question

How to find the endpoints of the common perpendicular between two skew lines in 3D

Original question: 7 12 The line $l_1$, which is parallel to the vector $2\mathbf{i}-4\mathbf{j}$, passes through the point $A$ whose position vector is $4\mathbf{i}+6\mathbf{j}-8\mathbf{k}$. The line $l_2$, which is parallel to the vector $\mathbf{i}+7\mathbf{j}-2\mathbf{k}$, passes through the point $B$ whose position vector is $5\mathbf{i}-6\mathbf{j}+12\mathbf{k}$. The point $P$ on $l_1$ and the point $Q$ on $l_2$ such that $\overrightarrow{PQ}$ is perpendicular to both $l_1$ and $l_2$. The point $H$ on the line segment $PQ$ such that $PQ = 4PH. Find the position vectors of $P$ and $Q$. [8] $l_1:\ \mathbf{r}=\begin{pmatrix}4\\6\\-8\end{pmatrix}+\lambda\begin{pmatrix}2\\-4\\0\end{pmatrix}$ $l_2:\ \mathbf{r}=\begin{pmatrix}5\\-6\\12\end{pmatrix}+\mu\begin{pmatrix}1\\7\\-2\end{pmatrix}$ $\overrightarrow{OA}=\begin{pmatrix}4\\6\\-8\end{pmatrix}$ $\overrightarrow{OB}=\begin{pmatrix}5\\-6\\12\end{pmatrix}$ $\overrightarrow{OP}=\begin{pmatrix}4+2\lambda\\6-4\lambda\\-8\end{pmatrix}$ $\overrightarrow{OQ}=\begin{pmatrix}5+\mu\\-6+7\mu\\12-2\mu\end{pmatrix}$ $\overrightarrow{PQ}=\begin{pmatrix}-1+\mu-2\lambda\\-12+7\mu+4\lambda\\20-2\mu\end{pmatrix}$ $\begin{pmatrix}-1+\mu-2\lambda\\-12+7\mu+4\lambda\\20-2\mu\end{pmatrix}\cdot\begin{pmatrix}2\\-4\\0\end{pmatrix}=0$ $\begin{pmatrix}-1+\mu-2\lambda\\-12+7\mu+4\lambda\\20-2\mu\end{pmatrix}\cdot\begin{pmatrix}1\\7\\-2\end{pmatrix}=0$ $-2+2\mu-4\lambda+48-28\mu-16\lambda=0$ $-10+\mu-2\lambda-84+49\mu+28\lambda-40+4\mu=0$ $-2\mu-20\lambda+46=0$ $53\mu+26\lambda-134=0$ $\mu-10\lambda+23=0$ $\lambda=0$ $\mu=-23$ $\lambda=-(-2)=2$ $\overrightarrow{OP}=\begin{pmatrix}4+2\\6-4(2)\\-8\end{pmatrix}=\begin{pmatrix}6\\-2\\-8\end{pmatrix}$ $\overrightarrow{OQ}=\begin{pmatrix}5-2\\-6-7(2)\\12-2(-2)\end{pmatrix}=\begin{pmatrix}3\\-20\\16\end{pmatrix}$

Expert Verified Solution

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Expert intro: This is a classic vector-geometry setup: two skew lines, a connecting segment that meets both at right angles, and a ratio condition that helps pin down the exact points. The clean way is to parameterize each line, write the vector PQ→\overrightarrow{PQ}, and use perpendicularity to build two equations for the parameters.

Detailed walkthrough

Let

r1=(46−8)+λ(2−40),r2=(5−612)+μ(17−2).\mathbf{r}_1=\begin{pmatrix}4\\6\\-8\end{pmatrix}+\lambda\begin{pmatrix}2\\-4\\0\end{pmatrix}, \qquad \mathbf{r}_2=\begin{pmatrix}5\\-6\\12\end{pmatrix}+\mu\begin{pmatrix}1\\7\\-2\end{pmatrix}.

So

P=(4+2λ, 6−4λ, −8),P=(4+2\lambda,\ 6-4\lambda,\ -8), Q=(5+μ, −6+7μ, 12−2μ).Q=(5+\mu,\ -6+7\mu,\ 12-2\mu).

Hence

PQ→=Q−P=(1+μ−2λ−12+7μ+4λ20−2μ).\overrightarrow{PQ}=Q-P=\begin{pmatrix} 1+\mu-2\lambda\\ -12+7\mu+4\lambda\\ 20-2\mu \end{pmatrix}.

Since PQPQ is perpendicular to both lines, we use dot products with the direction vectors.

1) Perpendicular to l1l_1

Direction of l1l_1 is (2,−4,0)(2,-4,0), so

PQ→⋅(2−40)=0.\overrightarrow{PQ}\cdot \begin{pmatrix}2\\-4\\0\end{pmatrix}=0.

That gives

2(1+μ−2λ)−4(−12+7μ+4λ)=02(1+\mu-2\lambda)-4(-12+7\mu+4\lambda)=0 2+2μ−4λ+48−28μ−16λ=02+2\mu-4\lambda+48-28\mu-16\lambda=0 26λ+13μ=25.26\lambda+13\mu=25.

2) Perpendicular to l2l_2

Direction of l2l_2 is (1,7,−2)(1,7,-2), so

PQ→⋅(17−2)=0.\overrightarrow{PQ}\cdot \begin{pmatrix}1\\7\\-2\end{pmatrix}=0.

Then

(1+μ−2λ)+7(−12+7μ+4λ)−2(20−2μ)=0(1+\mu-2\lambda)+7(-12+7\mu+4\lambda)-2(20-2\mu)=0 1+μ−2λ−84+49μ+28λ−40+4μ=01+\mu-2\lambda-84+49\mu+28\lambda-40+4\mu=0 26λ+27μ=123.26\lambda+27\mu=123.

Now solve

26λ+13μ=25,26\lambda+13\mu=25, 26λ+27μ=123.26\lambda+27\mu=123.

Subtracting the first from the second:

14μ=98⇒μ=7.14\mu=98 \Rightarrow \mu=7.

Then

26λ+13(7)=2526\lambda+13(7)=25 26λ=−6626\lambda=-66 λ=−3313.\lambda=-\frac{33}{13}.

So

P=(4+2(−3313), 6−4(−3313), −8)=(−1413, 30613, −8),P=\left(4+2\left(-\frac{33}{13}\right),\ 6-4\left(-\frac{33}{13}\right),\ -8\right) =\left(-\frac{14}{13},\ \frac{306}{13},\ -8\right),

and

Q=(5+7, −6+49, 12−14)=(12,43,−2).Q=(5+7,\ -6+49,\ 12-14)=(12,43,-2).

If your working uses the given ratio PQ=4PHPQ=4PH, that determines the point HH on segment PQPQ, but it is not needed to find PP and QQ once the perpendicular condition is used correctly.

💡 Pitfall guide

A common slip is to build PQ→\overrightarrow{PQ} with the wrong sign. It should be Q−PQ-P, not P−QP-Q, though either can work if you stay consistent. Another easy miss is forgetting that the third component of the first direction vector is 00, so the first dot-product equation is shorter than it looks.

🔄 Real-world variant

If the question instead asked for the foot of the perpendicular from one skew line to another, the process is the same: parameterize both lines, form PQ→\overrightarrow{PQ}, and enforce orthogonality to each direction vector. If a ratio like PQ=4PHPQ=4PH is meant to locate a point HH on the segment, then HH divides PQPQ in the ratio 1:31:3, so

H=P+14PQ→.H=P+\frac14\overrightarrow{PQ}.

That extra condition is useful when the midpoint or an internal division point is required.

🔍 Related terms

skew lines, vector equation of a line, perpendicular distance

FAQ

How do you find the common perpendicular points between two skew lines?

Write each line in vector form, let P and Q be the unknown points on the lines, form vector PQ, and set its dot product to zero with each line’s direction vector. Solving the two linear equations gives the parameters for P and Q.

What does it mean for PQ to be perpendicular to both lines?

It means the connecting segment PQ is orthogonal to each line’s direction vector. So PQ has zero dot product with both direction vectors.

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